Pascal’s law and hydraulic machines
Pascal’s law
Fill a plastic bottle with water and poke small holes all around it. Squeeze the bottle. Water squirts out of every hole, up, down and sideways, all with the same strength.
Blaise Pascal studied this in the 1600s. His first finding: at any point in a fluid at rest, the pressure is the same in every direction.
Here is NCERT’s proof. Inside the fluid, imagine a tiny wedge, a right-angled prism. Its sloping face has area A b, and its two flat faces have areas A c and A d.
The surrounding fluid pushes at right angles on each face, with forces F b, F c and F d. The wedge is so small that its weight can be ignored, and it is at rest, so these forces must balance.
Resolving the forces gives F b sin θ equals F c, and F b cos θ equals F d. The geometry of the wedge gives A b sin θ equals A c, and A b cos θ equals A d.
Divide each force by its area, and all three pressures come out equal: P b equals P c equals P d. Turn the wedge any way you like: the pressure at a point does not depend on direction.
A second fact: in a still fluid, all points at the same height have the same pressure. A horizontal cylinder of fluid is pushed equally from both ends, or it would start to move.
Now push on a piston that closes a container of liquid. Every point in the liquid feels the same extra pressure. This is Pascal’s law: a change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls.
Hydraulic lifts and brakes
In a car service station, a car weighing more than a tonne is lifted by a mechanic pressing a pedal. The trick is Pascal’s law.
A hydraulic lift has two cylinders joined by a pipe and filled with oil. A small piston of area A 1 is on one side, a large piston of area A 2 on the other.
Push the small piston down with force F 1. This adds a pressure P equals F 1 over A 1 to the oil, and Pascal’s law passes the same pressure on to the large piston.
The large piston feels a force F 2 equals P times A 2, which is F 1 times A 2 over A 1. If the large piston has fifty times the area, the force is fifty times bigger.
Worked example: the small piston has a radius of 5 centimetres, the large one 15 centimetres, and the car has a mass of 1350 kilograms. The car’s weight is 13 230 newtons.
The area ratio is 15 over 5, squared: 9. So the force needed is 13 230 divided by 9, about 1470 newtons. The pressure in the oil is about 1.9 times 10⁵ pascals, nearly twice atmospheric pressure.
Is this something for nothing? No. The oil does not compress, so the volume pushed out of the small cylinder equals the volume moved into the large one: A 1 d 1 equals A 2 d 2.
Push the small piston down by 9 centimetres, and the car rises only 1 centimetre. The force is multiplied, but the distance is divided. Work in, F 1 d 1, equals work out, F 2 d 2.
Hydraulic brakes in cars and motorcycles work the same way. Your foot pushes a small piston in the master cylinder, and the pressure travels through brake oil to larger pistons at the wheels.
Because the same pressure reaches every wheel, all of them are braked equally. A light push on the pedal becomes a large force on the brake pads. JCB excavators and dentists’ chairs use the same idea.
Try it: a hydraulic lift
- Press Push (or Space). The small piston moves 20 cm each time; watch how little the car rises.
- The pressure P = F₁/A₁ is the same everywhere in the oil.
- Lower F₁ below the “smallest F₁” readout: the car will not move.
- Make r₂ bigger. The force goes up as r₂², but the car rises less per push.
What is multiplied?
A hydraulic lift lets a small force raise a heavy car. Compared with the small piston, what is larger at the large piston?
- The pressure in the oil
- The force
- The work done
- The distance moved
Answer
B. The pressure is the same at both pistons (Pascal’s law), so the force is bigger on the bigger area: F₂ = F₁ × A₂/A₁. But the large piston moves a smaller distance, d₂ = d₁ × A₁/A₂, so the work done is the same on both sides. A machine cannot multiply energy.
Two syringes
Two syringes filled with water are joined by a tube. The small one has a piston of diameter 1.0 cm, the large one 3.0 cm. You push the small piston with a force of 10 N. What force does the water exert on the large piston?
- 3.3 N
- 10 N
- 30 N
- 90 N
Answer
D. Area goes as diameter squared, so A₂/A₁ = (3.0/1.0)² = 9. F₂ = 10 × 9 = 90 N. Using the diameter ratio instead of the area ratio gives 30 N, which is wrong. If you push the small piston in by 6 cm, the large one moves out only 6/9 ≈ 0.67 cm.
Summary
- Same pressure on both pistons:
F₁ / A₁ = F₂ / A₂ - Force multiplied by:
F₂ = F₁ × A₂ / A₁ - Oil is incompressible:
A₁ d₁ = A₂ d₂ - No free energy:
F₁ d₁ = F₂ d₂
- At a point in a fluid at rest, the pressure is the same in all directions. Points at the same height have the same pressure.
- Pascal’s law: a change in pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid and the walls.
- A hydraulic lift multiplies force by A₂/A₁, but the large piston moves a smaller distance, so the work in equals the work out.
- Hydraulic brakes pass the same pressure to all the wheels, so a light push on the pedal brakes every wheel equally.